<p>1- Last week the Star Bakery made 3 kinds of cakes. One half of the cakes were made with 4 eggs each, two thirds of the rest of the cakes were made with 3 eggs each, and the remaining 54 cakes were made with 2 eggs each. What was the total number of eggs used to make all of these cakes?</p>
<p>2- square tiles measuring 1/2 foot by 1/2 foot are sold in boxes containing 10 tiles each. What is the least number of boxes of tiles needed to cover a rectangular floor that has dimensions 12 feet by 13 feet?
(A) 7
(B) 16
(C) 32
(D) 57
(E) 63
3- in the xy-plane, line L passes through the point (4,-5) and the vertex of the parabola with the equation: y=-2(x-2)^2+3. What is the slope of the line L?
(A) -4
(B) -1/4
(C) 0
(D) 1/4
(E) 4</p>
<ol>
<li>Letting x be the number of cakes,
x/2 of the cakes had 4 eggs each
x/3 of the cakes had 3 eggs each (x/3 is “2/3 of the rest”).
x/6 of the cakes had 2 eggs each (x/6 is the remaining part).</li>
</ol>
<p>The question implies that x/6 = 54 → x = 324. Now it’s just arithmetic, which I’ll leave up to you.</p>
<ol>
<li><p>First we find the number of tiles. Since we can simply tile the whole floor without overlaps, the number of tiles is just (12<em>13)/(.5</em>.5) = 624 tiles. 624/10 = 62.4, but we round up to 63, E.</p></li>
<li><p>The vertex of the parabola occurs at x = 2, y(2) = 3. So line L passes through (2,3) and (4,-5). Find the slope.</p></li>
</ol>
<p>@rspence, thank you for your time and response, but I have 2 problems:
1- How do I know the vertex of a parabola?
2- I did not quite understand the explanation for the first question. Could you please clarify it a bit more?</p>
<p>The following is right from my article “The Math Formulas You Should Memorize for the SAT.” </p>
<p>The general form for a quadratic function is y = ax2 + bx + c. </p>
<p>The graph of this function is a parabola whose vertex has x-coordinate x = -b/2a</p>
<p>The parabola opens upwards if a > 0 and downwards if a < 0.</p>
<p>Here is an example.</p>
<p>Let the function f be defined by f(x) = -2x^2 - 3x + 2. For what value of x will the function f have its maximum value?</p>
<p>The graph of this function is a downward facing parabola, and we see that a = -2, and b = -3. So the x-coordinate of the vertex is x = 3/(-4) = -3/4 .</p>
<p>The standard form for a quadratic function is y - k = a(x-h)^2 </p>
<p>The graph is a parabola with vertex at (h,k). Again, the parabola opens upwards if a > 0 and downwards if a < 0.</p>
<p>Here is an example.</p>
<p>Let the function f be defined by f(x) = 3(x - 1)2 + 2. For what value of x will the function f have its minimum value?</p>
<p>The graph of this function is an upward facing parabola with vertex (1,2). Therefore, the answer is 1.</p>
<p>Remark: Note that in this example k = 2, and it is on the right hand side of the equation instead of on the left.</p>
<p>The vertex of a parabola in the form y = ax^2 + bx + c occurs at x = -b/2a (this is easily proved using calculus by taking the derivative of both sides).</p>
<p>If the parabola in the form y = a(x-h)^2 + k, the vertex is at (h,k). Clearly, this must have some equivalence with the first statement, which it does (h = -b/2a). So the vertex of the parabola y = -2(x-2)^2 + 3 is (2,3).</p>
<p>Also, 1/2 of the cakes had 4 eggs each. How much is two thirds of the rest? Here, “the rest” is simply the remaining 1/2, and two thirds of 1/2 is 1/3. So 1/3 of the cakes had 3 eggs each. The remaining 54 cakes had two eggs each, 1 - 1/2 - 1/3 = 1/6, so 54 cakes comprises 1/6 of the cakes.</p>
<p>about vertex of a parabola…
y=ax²+bx+c (a≠0)
=a[x²+(b/a)x+(b/2a)²-(b/2a)²+(c/a)]
=a[(x+b/2a)²-(b/2a)²+(c/a)]
=a(x+b/2a)²-(b²-4ac)/4a
when x=-b/2a, y=(4ac-b²)/4a (max/min)</p>
<p><em>Optional</em> Another derivation of the vertex of a parabola.</p>
<p>y = ax^2 + bx + c. In calculus there is a certain operator called a “derivative” that basically tells you how fast a function value is changing with respect to y. For 2D functions, this is equal to the slope of a tangent line through that point (for continuous functions) and is usually denoted dy/dx for this problem (some texts use y’, other notations as well).</p>
<p>For this parabola, dy/dx = 2ax + b. The vertex of the parabola occurs when the slope of the tangent line at a point (x0, y0) is zero. If we set dy/dx = 2ax + b = 0, we obtain one solution, x = -b/2a, which turns out to be the x-coordinate of the vertex of the parabola.</p>
<p>The advantage of knowing the derivative is that this works for higher-degree polynomials, as well as any differentiable function in general. But you won’t need this for the SAT…</p>
<p>A non-calculus derivation of the vertex
let f(x)=ax^2+bx+c. We wish to see if a value k exists such that f(k+x)=f(k-x) for all x. If the graph has symmetry about k, then k must be either a max or min. Doing some algebra we get
a(x+k)^2+b(x+k)+c=a(k-x)^2+b(k-x)+c, cancelling c’s and expanding:
ax^2+2akx+ak^2+bx+bk=ak^2-2akx+ax^2+bk-bx. We can cancel ax^2,ak^2, and bk
leaving us with 2akx+bx=-2akx-bx, 4akx=-2bx, k=(-2bx)/(4ax) so k=-b/(2a), the result rspence got using calculus.</p>
<ol>
<li>If you know what a vertex is, i.e the lowest or highest point, then you should know that what you’re looking for is a highest or lowest value for Y. For this sort of a value to exist there must be some sort of a limit set, and these are mostly set by what is being added or subtracted. Note that 3 is being added in the equation, so now you just need to figure out if there is a highest or lowest for the other part. X-2 becomes 0 if you put X=2, and thus that whole part becomes 0 (and you get y=3). Then to make sure this is the value check if you can get a lower value if you use a different value of X, a highest since the whole variable part if being squared it is impossible for the that side of the equation to equal a negative number, thus it can be assumed that whatever value of X is given the value of Y will not drop below Y.</li>
</ol>
<p>For the record, I have never seen an SAT problem that required a student to find the vertex of a parabola in standard form. So if you know that x=-b/2a it can’t hurt but is not necessary. Same thing for the calculus.</p>
<p>What you DO want to be able to do is look at a parabola such as: y = -2(x-2)^2 + 3</p>
<p>and see that compared to a basic y=x^2 parabola…</p>
<p>i. it is “upside down” because of the -2 in front
ii. it has been shifted two units to the right because of the (x-2)
iii. it has been shifted up 3.</p>
<p>So the vertex is a (2,3 )</p>
<p>If anyone has an example of a college board SAT problem that requires you to find the vertex by using -b/2a, calculus, completing the square, etc, I’d like to see it.</p>
<p>i found the easiest way to do it is factor -2(x-2)^2+3 out and get -2x^2+8x-5. plug into graphing calculator and you get a downward facing parabola with vertex (2,3) simply plug in the formula rise/run=slope and you get (3+5)/(2-4) (y1-y2)/(x1-x2) you get the answer as a the slope of line l = -4. hope that helps</p>
<ol>
<li>is quite simple the area of the rectangular floor is length x width. 12 x 13 = 156 feet if each tile is 1/2 sq foot then double the area is needed in tiles (set up a proportion if you’re unsure) multiply 156 x 2= number of tiles, then divide by the amount per package (10) to receive the number of packages (32)</li>
</ol>
<p>in #1 you are left with a remainder that is a fraction of the whole. 1/2 of the total is 4 eggs, 2/3 of the remaining half is 3 eggs, and the remainder of the rest is 54 cakes with 2 eggs. i split the fraction into 6ths and was left with 3/6 of total is 4, 2/6 of total is 3, 1/6 of total =54. multiply 54 x 6 to receive the total number of eggs. after that the answer is arithmetic (162 x 4) + (108 x 3) + (54 x 2)=1080 eggs total</p>